I am trying to solve a puzzle through solving a series of questions, here is the first version of the puzzle:

Is it possible to find 2 connected subsets $A$ and $B$ inside the square $I \times I$ such that $(0,0),(1,1) \in A$ and $(0,1),(1,0) \in B$ and $A \cap B = \emptyset.$

We know that the answer is YES. But we are guessing that the answer is NO if we required $A$ and $B$ to be path connected.Our intuition comes from the attitude of the Topologist's sine curve. Now we are trying to write a proof of this guess. During this path we wanted to prove the Jordan Curve Theorem. So I am trying to do so.

Here are the questions that I should have solved so far:

Proving the equivalence of the Criss - Cross theorem statement(ordinary one) to another statement.

Proving the criss cross theorem (generalized version). (I did not solve this question yet, so any help with the solution will be appreciated )

Also I know the proof of the ordinary version of the criss-cross theorem. Here is it:

Assume towards contradiction that we have such paths $\alpha$ and $\beta$ such that $\alpha (s) \neq \beta (t)$ for all $s,t \in I.$ This can be written as $\alpha (s)- \beta (t) \neq 0$ for all $s,t \in I.$ Which can further be rephrased by defining $H: I \times I \rightarrow \mathbb{R}^2$ by the formula $$H(s,t) = \alpha(s) - \beta (t).$$ And our assumption means that the function $H$ never takes the value $0 \in \mathbb{R}^2;$ in diagram language, we assume that there is a continuous function $\hat{H}$ making the following diagram commutative:

$$\require{AMScd} \begin{CD} I \times I @>{\hat{H}}>> \mathbb{R}^2 - \{0\}\\ @VVV @VVV \\ I\times I @>{H}>> \mathbb{R}^2 \end{CD}$$

where the arrow below $\hat{H}$ should be a dotted arrow because we are searching for this function. And I am not skillful in drawing commutative diagrams this is why I draw $I \times I$ 2 times because I do not know how to draw one dotted arrow coming out of $I \times I$ going directly to $\mathbb{R} - \{0\}$ my bad. Then my job is to show that there can be no such function $\hat{H}.$

Now, I want to solve this problem:

Suppose there were such a function $\hat{H},$ and write $\hat{h}$ for the restriction of $\hat{H}$ to the boundary of $I \times I.$ Sketch what the image of $\hat{h}$ might look like - remember that there is no guarantee or assumption that either $\hat{H}$ or $\hat{h}$ is injective, so put some possible self-crossing in your sketch.

I do not know how to sketch it. could anyone help me in doing so please?


The key observation is that you know a little bit about what $\hat{h}$ looks like.

Assume $\alpha : (0,0)\to (1,1)$ and $\beta : (0,1)\to (1,0)$ are your paths.

Then $$\hat{h}(0,0) = \alpha(0)-\beta(0) = (0,0)-(0,1) = (0,-1).$$ as $t$ runs from $0$ to $1$, we get $$\hat{h}(t,0) = \alpha(t)-\beta(0) = (q,r) - (0,1) = (q,r-1).$$ We know almost nothing about this path except that $0\le q\le 1$ and $-1\le r\le 0$ and $(q,r)\ne (0,1)$, so drawing any path that stays inside this square and doesn't touch the origin is acceptable.

Then $\hat{h}(1,0) = (1,1)-(0,1) = (1,0)$. For $0\le s\le 1$, $\hat{h}(1,s)$ stays in $[0,1]\times[0,1]\setminus\{(0,0)\}.$ Again, drawing any path that stays in here is acceptable. Next $\hat{h}(1,1) = (1,1)-(1,0) = (0,1)$, and as $t$ goes from $1$ back to $0$ on the next segment, $\hat{h}(t,1)$ stays inside $[-1,0]\times [0,1]$ and doesn't touch the origin.

Finally, $\hat{h}(0,1) = (-1,0)$, and as $s$ returns to $0$, $\hat{h}(0,s)$ stays inside $[-1,0]\times[-1,0]$ and doesn't touch the origin before returning to $(0,-1)$.

You'll notice that any path you try to draw that satisfies these requirements forms a counterclockwise loop about the origin that is nonzero in the fundamental group, contradicting the assumption of the existence of the extension $\hat{H}$.

Note that this proves the sets in your puzzle cannot exist if they are path connected. Not sure why you would need the Jordan curve theorem.

  • $\begingroup$ Maybe at the end of my question # 16 I think it will be clear why we would need Jordan curve theorem. $\endgroup$ – user778657 Apr 27 '20 at 2:23
  • $\begingroup$ what do you mean by "nonzero in the fundamental group"? and how this will contradict the assumption of the existence of the extension $\hat{H}$? $\endgroup$ – user778657 Apr 28 '20 at 2:46
  • $\begingroup$ @Smart20 I mean not contractible in the ambient space, and if $\hat{H}$ existed, it would give a contraction of the loop. $\endgroup$ – jgon Apr 28 '20 at 3:01

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